 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/17/c/solution

u = x^{k}

and

d v = sin x d x

F (k) = [- x^{k} cos x]_{0}^{π} + k \int_{0}^{π} x^{k - 1} cos x d x = π^{k} + k \int_{0}^{π} x^{k - 1} cos x d x .

For

k > 1

, integrate the remaining integral by parts once more. Its boundary term

[x^{k - 1} sin x]_{0}^{π}

is zero, so

F (k) = π^{k} - k (k - 1) F (k - 2) .

k = 1

, direct integration gives

F (1) = π

, agreeing with the same formula; the coefficient of

F (- 1)

vanishes and that integral is finite. Now follow the odd chain:

F (3) = π^{3} - 6 π, F (5) = π^{5} - 20 π^{3} + 120 π .

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